Math 30-2 Diploma common mistakes
Math 30-2 Diploma · Understand the traps and how to avoid them.
When a question asks for the number of arrangements and gives a condition like 'the two vowels must be together', the candidate counts the two restricted items as separate positions and multiplies by 2! but forgets to treat the glued block as one item when dividing out repeated letters, producing an answer that is off by a factor of the repeated-letter count.
The block method is taught as a single trick ('glue, arrange, multiply by internal arrangements'), so students apply the glue-and-multiply steps but never revisit the repeated-element division they did in the unrestricted part of the problem.
What to do instead
Write the counting plan in two labelled lines before computing: Line 1 'items to arrange = (n-1) blocks + internal 2!'; Line 2 'divide by factorials of any repeated items still inside the blocks'. Recompute the repeated-letter count after gluing, since letters inside the glued block no longer count as separate positions.
On a 'determine the validity of the reasoning' item, the candidate judges the conclusion by whether the conclusion itself is true in real life rather than by whether it follows from the premises, so a valid argument with a false premise is marked invalid or an invalid argument with a true conclusion is marked valid.
Everyday reasoning rewards true conclusions, and the exam's artificial premises (e.g., 'all fish can fly') feel wrong, so truth-value intuition overrides the formal definition of validity.
What to do instead
Adopt a two-column habit: left column 'given (accept as true, even if absurd)', right column 'what must follow'. Cover the conclusion, derive what is forced, then uncover and compare. Only after deciding validity, ask separately whether the premises are actually true.
The candidate confuses the converse and inverse of a conditional with its contrapositive, treating 'if not P then not Q' as logically equivalent to 'if P then Q' and selecting it as a valid deduction.
All three transformed statements look like rearrangements of the same sentence, and only the contrapositive preserves truth value, which is a subtle distinction that is easy to blur under time pressure.
What to do instead
For each conditional, mechanically write all three transforms with labels: converse (Q→P), inverse (¬P→¬Q), contrapositive (¬Q→¬P). Circle only the contrapositive as equivalent. Test with a concrete counterexample sentence (e.g., 'if it is a dog then it is an animal') to confirm the inverse fails.
When computing the probability of 'at least one' success across repeated independent trials, the candidate adds the individual probabilities (e.g., 0.3 + 0.3 + 0.3) instead of using the complement, yielding a probability greater than what is possible or ignoring overlap.
'At least one' sounds additive, and the complement method requires an extra conceptual step (find P(none), then subtract from 1) that is not the most obvious operation.
What to do instead
Trigger a fixed routine whenever the words 'at least one' appear: compute P(none) = (1-p)^n, then answer 1 - P(none). Sanity-check that any 'at least one' answer lies between the single-trial probability and 1.
The candidate computes a conditional probability P(A|B) by dividing by the total sample space instead of by P(B), or reverses the condition and reports P(B|A) when P(A|B) was asked.
The notation P(A|B) is read left-to-right as 'A given B' but students often process the bar as a fraction separator and divide by the wrong total, and the distinction between the two conditional directions is invisible in a word problem unless carefully parsed.
What to do instead
Before calculating, rewrite the question in the form 'restrict to the group that ___; within that group, what fraction ___'. The first blank is the denominator's condition. For two-way tables, physically cover the rows or columns that do not meet the condition before counting.
When a question involves selecting a committee or handshake-type grouping, the candidate uses permutations (order matters) instead of combinations, or multiplies probabilities for dependent selections as if the items were replaced.
The words 'choose' and 'select' feel neutral, and the order/no-order distinction is not signalled by any keyword, so students default to whichever formula they practised most recently.
What to do instead
Ask one test question: 'If I swapped two selected items, would that be a different outcome?' If no, use combinations. For multi-stage selections without replacement, decrease both the numerator and the denominator by one after each draw and write the shrinking denominator explicitly.
When given a graph or table of a relation and asked for the domain or range, the candidate lists the x-values for the range (or y-values for the domain), or omits repeated values and endpoints inconsistently, producing a set that is reversed or incomplete.
Domain and range are defined by a single letter each (x vs y) that is easy to swap under recall pressure, and students often read the table top-to-bottom rather than axis-by-axis.
What to do instead
Annotate the axes before answering: write 'domain = x = left-right' and 'range = y = up-down' directly on the paper. Read the graph by sweeping a vertical line for domain and a horizontal line for range, recording where the relation exists.
When a function is given in the form y = af(b(x-h)) + k, the candidate applies the horizontal stretch or reflection in the wrong order relative to the horizontal shift, or shifts by h in the wrong direction, so the transformed graph is displaced incorrectly.
Horizontal transformations compose in the opposite order to the way they are written and to the way vertical transformations behave, and the inside-the-bracket operations feel symmetric with the outside ones.
What to do instead
Factor the inside of the bracket to the form b(x - h) first, so the shift value and direction are explicit. Then apply transformations in the order: horizontal stretch/reflection about the y-axis, then horizontal translation; separately vertical stretch/reflection, then vertical translation. Track one key point through each step.
When solving a rational equation, the candidate multiplies both sides by the common denominator but forgets to state or check the restriction that the denominator cannot be zero, and reports a solution that makes an original denominator vanish.
The restriction is a side condition, not part of the algebraic manipulation, so once the equation is cleared of fractions the excluded value is no longer visible.
What to do instead
As the very first line of work, list the non-permissible values by setting each original denominator to zero. After solving, compare every candidate root against that list and reject matches, stating the rejection explicitly.
When solving a trigonometric equation such as sin θ = 0.5 over a stated interval, the candidate reports only the calculator's principal answer and misses the second solution in the other quadrant, or gives answers in the wrong unit because the calculator is in the wrong mode.
Inverse-trig functions return a single value, and the symmetry that produces the second solution within 0° to 360° is not visible from the calculator display; degree/radian mode is a silent setting.
What to do instead
Check the calculator mode before any trig entry. Then use a reference-angle template: find the reference angle, and construct all solutions in the required interval using the CAST/quadrant pattern (for sine, also 180° - reference). List every solution and verify each falls inside the stated interval.